Question

A proton, a deutron and an $\alpha$-particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to it.Ratio of radii of circular paths respectively

• A. $1:1:2$
• B. $\sqrt{2}:\sqrt{2}:1$
• C. $1:2:1$
• D. $2:1:1$

Answer 1

The correct option is B $\sqrt{2}:\sqrt{2}:1$

The radius of trajectory of the particle

$r=MV/BQv=\sqrt{2VQ}/\sqrt{M}$ where V is the acc.potential

$r=M^{1/2}\sqrt{2}V/\sqrt{Q}B$

Hence

$r_{proton}=M{e}^{1/2}\sqrt{2}V/\sqrt{e}B{r}_{duetron}=\sqrt{2}M{e}^{1/2}\sqrt{2}V/\sqrt{e}{r}_{alpha}=2M{e}^{1/2}\sqrt{2}V/\sqrt{2}\sqrt{e}$

Hence ratio$=1:\sqrt{2}:\sqrt{2}$